2iz / (1-i) + 2i = 2z (1-i) -3 + i

Let z = a+ bi

==> 2i(a+bi)/ (1-i) + 2i = 2(a+bi)(1-i) -3 + i

Multiply by (1-i):

==> 2i(a+bi) + 2i(1-i) = 2(a+bi)(1-i)(1-i) -3(1-i)+ i(1-i)

==> 2ai - 2b + 2i +2 = 2(a+bi)(-2i) -3 +3i + i +1

==> 2ai -2b + 2i +2 = -4ai +4b -3 +3i +i +1

Now group similar terms:

==> 2ai -2b +4ai -4b = -3+3i +i +1 -2i -2

==> -6b + 6ai = -4 +2i

==> -6b = -4 ==> b= 4/6= 2/3

==> 6a = 2 ==> a= 2/6= 1/3

==> z= (1/3) + (2/3)i

We'll note z = a + b*i and we'll substitute z into the given equation.

2i*(a + b*i)/(1-i) + 2i = 2(a + b*i)(1-i) - 3 + i

We'll multiply the terms 2i, 2(a + b*i)(1-i), - 3 and i by (1-i):

2i*(a + b*i) + 2i*(1-i) = 2(a + b*i)(1-i)^2 - 3(1-i) + i*(1-i)

We'll remove the brackets:

2ai - 2b + 2i + 2 = -4ai + 4b - 3 + 3i + i + 1

We'll move all the terms in a and b to the left side and all the terms without a and b, to the right side:

2ai - 2b + 4ai - 4b = -2i - 2 - 3 + 3i + i + 1

We'll combine the real parts and the imaginary parts:

-6b + 6ai = -4 + 2i

Because the expressions from both sides are equivalent, the real parts and the imaginary parts have to be equal.

-6b = -4

**b = 2/3**

6a = 2

**a = 1/3**

**z = 1/3 + 2i/3**

2iz/(1-i) +2i = 2z(1-i)-3+i

Solution:

We realise the denominator in 2iz/(1-i) = 2iz(1+i)/(1^2-i^2 ) = 2z((i+i^2)/(i-i^2) = 2z(i-1)/2 = z(i-1).

So the equation now becomes:

2z(i-1) +2i = 2z(1-i) -3+i. Colect similars.

z (i-1 )-2z(1-i) = -3+i-2i

z( i-1 -2+i) = -3-i

z(-3) = -(3+i)

z = -(3+i)/(-3) = 1 + (1/3)i

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